**Definition: **Let G be a group under a binary operation * and H non empty subset of G. Then H is said to be a subgroup of G if H itself forms a group under the same binary operation * in G.

Theorem: A non empty subset H of a group G, is a subgroup of G if and only if a, b ∊ H ⟹ ab^{-1} ∊H

**Proof:** Let H be a subgroup of G. Then H itself forms a group under the operation in G. Hence by the closure and the inverse properties in H, a, b ∊H ⟹a, b^{-1} ∊H ⟹ab^{-1}∊H

Conversely let a, b ∊H ⟹ab ∊H

Then a ∊H ⟹ a a^{-1} = e∊H, where e is the identity element of G. Also e, b∊H ⟹eb^{-1}=b^{-1}∊H .

Thus a, b∊H ⟹a(b^{-1})^{-1} = ab∊H .

Also assertively holds for H, as it does for G. It is observed that H satisfies the defining proprieties of a group and so is a subgroup of G. ** (Proved) **

Theorem: Let G be a group and H a nonempty finite subset of G. Then G is a subgroup of G if and only if a, b ∊H ⟹ab ∊H

**Proof:** Let H be a subgroup of G. Then H must be closed under the operation in G, i.e., a, b ∊H ⟹ ab ∊H .

Conversely let a, b ∊H ⟹ ab ∊H.

Then for any a∊H, we have a^{2} = a . a ∊H, a^{3} = a^{2} a ∊H, a^{4} = a^{3} a ∊H, …, a^{n} ∊H, where n is any positive integer.

Since H is finite, the elements a, a^{2}, a^{3},…, a^{n} … in H cannot all be distinct. This means there exist positive integers r, s (r > s)

Such that a^{r} = a^{s} ——————-(i)

Applying the cancellation law in G, to (i)

We have a^{r-s} = e, the identity element of G.

Since r – s>0, a^{r-s} = e ∊ H.

Also r – s -1 ≥ 0 and since a^{0} = e ∀ a ∊ G,

We have a^{r-s-1} ∊ H. Again, a a^{r-s-1} = a^{r-s} =e

⟹ a^{r-s-1} is the inverse a in G.

⟹ a^{r-s-1 }= a^{-1} ⟹ a^{-1} ∊ H.

*Hence H is a subgroup of G. (Proved)*